similar to: How to create following chart for visualizing multivariate time series

Thanks David, It is working. Holtman's also gave me a solution but, I wanted to have a color pallet for description of colors, that was not in his solution. However I need one small modification. If I want to plot only lower diagonal elements of 'dat' then how should I proceed? What I want is, to visualize only lower diagonal elements and having the color pallet on them only. Also

Hi all, Can anyone here please tell me whether is it possible to produce a chart displayed in http://www.datawolf.blogspot.com/ in R for visualizing multivariate time series? If possible how? Regards, --------------------------------- [[alternative HTML version deleted]]

I used ?image function to do that, like below : require(grDevices) # for colours x <- y <- seq(-4*pi, 4*pi, len=27) r <- sqrt(outer(x^2, y^2, "+")) image(x, y, r, col=gray((0:32)/32)) However my next problem to add a color pallet for color description [as shown in following link]. If anyone here tell me how to do that, it will be good for me. Regards, Megh Dal

hi, hi, good morning everyone. I have a time series with binary outputs like : 0001011110100.................etc. Now I want to forecast the future values of that. Can anyone please tell me whether there is any tools exist in literature for dealing with this kind of binary observation? If possible please provide me some good references in net as well. rgd, Megh

Dear all R users, I have a small doubt about panel data analysis. My basic understanding on Panel data is a type of data that is collected over time and subjects. Vector Autoregressive Model (VAR) model used on this type of data. Therefore can I say that, one of statistical tools used for analysis of panel data is VAR model? If you clarify my doubt I will be very grateful. Thanks and regards,

Dear all R users, Please forgive me if my question is too trivial. Suppose I have two variables, (x,y) which is log-normally distributed with expected value (mu1, mu2) and some variance-covariance matrix. Now I want to draw a random sample of size 1000 from this distribution. Is there any function available to do this? Thanks and regards, Megh

The min_height = -0.25 is there to make it show cycle values down to -1/4. You may want to change it to -1 so it shows more of the cycle values. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Oct 17, 2017 at 1:26 PM, Roy Mendelssohn - NOAA Federal < roy.mendelssohn at noaa.gov> wrote: > yes, thanks, and I was getting close to that. One thing I found is the > manual says the

Thanks again. I am going to try the different versions. But I probably won't be able to get to it till next week. This is probably at the point where anything further should be sent to me privately. -Roy > On Jun 1, 2017, at 1:56 PM, David L Carlson <dcarlson at tamu.edu> wrote: > > On the off chance that anyone is still interested, here is the corrected function using

yes, thanks, and I was getting close to that. One thing I found is the manual says the height is the distance above the y-line, which should be, but doesn't have to be positive. In fact, the time series are estimates of a cycle, and has negative values, which unfortunately are not included in my sub-sample. And the negative values are not handled properly (the series disappears for

On the off chance that anyone is still interested, here is the corrected function using aperm(): z <- array(1:120,dim=2:5) f2 <- function(a, wh) { idx <- seq_len(length(dim(a))) dims <- setdiff(idx, wh) idx <- append(idx[-1], idx[1], wh-1) aperm(apply(a, dims, rev), idx) } all.equal(f(z, 1), f2(z, 1)) # [1] TRUE all.equal(f(z, 2), f2(z, 2)) # [1] TRUE

Does the following work for you? ggplot2::ggplot(plotFrame, aes(x = time, y = depth, height = cycle, group = depth)) + ggridges::geom_ridgeline(fill="red", min_height=-0.25) Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Oct 17, 2017 at 12:43 PM, Roy Mendelssohn - NOAA Federal < roy.mendelssohn at noaa.gov> wrote: > I have tried: > > ggplot(plotFrame, aes(x =

Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, >>> junk1 <- junk[, rev(seq_len(10), ] so that junk[1,1,1 ] = junk1[1,10,1] junk[1,2,1] = junk1[1,9,1] etc. The genesis of this is the program is downloading data from a variety of sources on (time, altitude, lat, lon) coordinates, but all

I have 2 vecros : x<-c(100,96,88,100,100,96,80,68,92,96,88,92,68,84,84,88,72,88,72,88) x1 = sample(x, 5, replace=FALSE) Now i want to get remaining values of vector "x" those are not member of vector "x1". Can anyone please tell me how to do that?

Thanks to all that replied. I had just looked through the httr code and sure enough for a .csv mime time it calls readr::read_csv(). The httr::content docs suggest not using automatic parsing in a package, rather to determine mime type and parse yourself and Ben's suggestion also works if I do: junk <- readr::read_csv(r1$content, col_types = cols()) Perfect. Using httr rather than

> On 1 Jun 2017, at 22:42, Roy Mendelssohn - NOAA Federal <roy.mendelssohn at noaa.gov> wrote: > > Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave, > >>>> junk1 <- junk[, rev(seq_len(10), ] > > > so that > > junk[1,1,1 ] = junk1[1,10,1] > junk[1,2,1] =

I have tried: ggplot(plotFrame, aes(x = time, y = cycle, height = cycle, group = depth)) + geom_ridgeline() ggplot(plotFrame, aes(x = time, y = depth, height = cycle, group = depth)) + geom_ridgeline() ggplot(plotFrame, aes(x = time, y = depth, group = depth)) + geom_density_ridges() none are producing a plot that was a ridgeline for each depth showing the time series at that depth. The plot

My error. Clearly I did not do enough testing. z <- array(1:24,dim=2:4) > all.equal(f(z,1),f2(z,1)) [1] TRUE > all.equal(f(z,2),f2(z,2)) [1] TRUE > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" # Your earlier example > z <- array(1:120, dim=2:5) >

Although it is not an elegant solution, but if your output format is HTML, you can add an arbitrary empty HTML element like <span id="foo"></span> before your code chunk. Then you can jump to this <span> via a link like "see [this code chunk](#foo)". Regards, Yihui -- https://yihui.name On Tue, Aug 29, 2017 at 1:30 PM, Roy Mendelssohn - NOAA Federal

How about this: f <- function(a,wh){ ## a is the array; wh is the index to be reversed l<- lapply(dim(a),seq_len) l[[wh]]<- rev(l[[wh]]) do.call(`[`,c(list(a),l)) } ## test z <- array(1:120,dim=2:5) ## I omit the printouts f(z,2) f(z,3) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into

?? > z <- array(1:24,dim=2:4) > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" In fact, > dim(f(z,3)) [1] 2 3 4 > dim(f2(z,3)) [1] 3 4 2 Have I made some sort of stupid error here? Or have I misunderstood what was wanted? Cheers, Bert Bert Gunter